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Q.

e2x+logxdx=

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a

12(2x1)e2x+c

b

14(2x1)e2x+c

c

14(2x+1)e2x+c

d

12(2x+1)e2x+c

answer is A.

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e2x+logxdx=xe2xdx=xe2x212e2xdx+c=e2x4(2x1)+c.

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